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Current Question (ID: 7511)

Question:

$\text{The mass ratio of photons corresponding to the first lines of the Lyman and the Balmer series of the atomic spectrum of hydrogen is :}$

Options:

1. $5 : 27$
2. $1 : 4$
3. $27 : 5$
4. $4 : 1$

Solution:

$\text{Hint: Use } \Delta E = mc^2$\n\n$\text{Step 1:}$\n\n$\Delta E = mc^2 \text{ ....... 1}$\n\n$\Delta E = hcz^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ ..... 2}$\n\n$\text{Compare equation 1 and 2 we get,}$\n\n$mc^2 \propto \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$\n\n$\text{For Lyman series, first-line obtained when the transition occurs between energy level 1 to 2}$\n\n$\text{For Balmer series, first-line obtained when the transition occurs between energy level 2 to 3}$\n\n$\text{Now compare the two equations:}$\n\n$\text{For Lyman series, } m_1c^2 \propto \left(\frac{1}{1^2} - \frac{1}{2^2}\right)$\n\n$\text{For Balmer series, } m_2c^2 \propto \left(\frac{1}{2^2} - \frac{1}{3^2}\right)$\n\n$\frac{m_1}{m_2} = \frac{(\frac{3}{4})}{(\frac{5}{36})} = \frac{27}{5} = m_1:m_2 = 27:5$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}